Probability distributions: True False questions
True false questions
Below you can find a bunch of simple questions that are either true or false. The questions are based on Chapters 6 to 10 from Introduction to Probability by Joseph Blitzstein and Jessica Hwang.1Below I often abbreviate this to BH. The questions are meant to help you practice with the definitions and some of the theorems discussed in the book. They are not meant as substitution for the problems of the book itself.
1. Week 1
\(X\) is a rv, \(X \in \R\). Claim: \(\supp{X} = (c, \infty) \implies \P{X\leq c} = 0\).
Solution
Solution, for real
True.
Suppose \(X\) is a real-valued rv with \(\supp{X} = [0, c]\). Claim:
\begin{equation*} \V{X} = \E{X^2} - (\E{X})^2 \leq c \E{X} - (\E{X})^2 = (c-\E{X})\E{X}. \end{equation*}Solution
Solution, for real
True.
Let \(X\) and \(Y\) be independent rvs. Claim: \(F_{X+Y}(x, y) = F_X(x) + F_{Y}(y)\).
Solution
Solution, for real
False. We should write \(F_{X,Y}\) rather than \(F_{X+Y}\), and since the rvs are independent, the joint CDF is the product of the marginal CDFs, not their sum.
We have two positive rvs \(X\) and \(Y\). Claim: \(\V{X+Y} = \V{X} + \V{Y}\).
Solution
Solution, for real
False. It’s not given that \(X\) and \(Y\) are independent.
Write \(m\) for the median of the rv \(X\). Claim: the following definition is correct:
\begin{equation*} \V{X} := \E{X^2} - m^{2}. \end{equation*}Solution
Solution, for real
False. \(\E{X}\) need not be equal to the median \(m\), and the definition of the variance involves the mean, not the median.
Let \(A\) and \(B\) be two events. Claim: \(\E{\1{A}\1{B}} = \P{A} + \P{B} - \P{A \cup B}\).
Solution
Solution, for real
True.
An unfair 4-sided die throws 4 half of the time and 1 to 3 with equal probability. Let the rv \(X\) denote the thrown value of the die. Claim: \(\E{X^2} = 38/3\).
Solution
Solution, for real
False. Calculating using LOTUS gives \(\E{X^2} = \frac{1}{2} 4^{2} + \frac{1}{2\cdot3}(1 + 4 + 9) = \frac{31}{3}\).
Let \(X\) be a degenerate rv and \(c\) a non-zero constant. Claim: \(\V{cX} > 0\).
Solution
Solution, for real
False. The variance of a degenerate rv is always 0. See the warning in BH.4.1.3 in which degeneracy is discussed.
Assume that \(\V{X} = \sigma^2\) and \(\E{X^2} = a^2\) both exist and are finite. Claim: \(\E{X} = \sqrt{a^2-\sigma^2}\).
Solution
Solution, for real
False. Because \(X\) can be negative.
Suppose \(\V{X+Y} = \V{X} + \V{Y}\). Claim: \(X\) and \(Y\) are independent.
Solution
Solution, for real
False. Independence is sufficient but not necessary for the equality to hold.
Let \(X\) and \(Y\) be two rvs where \(Y\) is always equal to \(X\), and \(\V{X}>0\). Claim: \(\V{X+Y} = \V{X} + \V{Y}\).
Solution
Solution, for real
False. Since \(Y\) is always equal to \(X\), they are definitely not independent. (see BH. p. 172) In fact, as \(Y=X\), \(\V{X+Y} = \V{2X} = 4 \V{X}\). Isn’t it a bit counterintuitive that when \(X\) and \(Y\) are dependent like this, the variance is larger than if they were independent?
Claim: The expectation of a continuous rv must always be nonnegative given that the probability density function values are nonnegative.
Solution
Solution, for real
False. The density values are nonnegative. The values that the RV can attain can be negative, therefore the expectation may be negative.
Let \(X \sim \Norm{0, 1}\). Claim: \(\P{X = 0} > \P{X = 5}\).
Solution
Solution, for real
False. By definition \(\P{X = k} = 0 \quad \forall k\).
Let \(U \sim \Unif{a,b}\) and \((c,d) \subset (a,b)\). Claim: the conditional distribution of \(U\) given that \(U \in (c,d)\) is \(\Unif{c,d}\).
Solution
Solution, for real
True.
See BH proposition 5.2.3:
Proof. For \(u\) in \((c, d)\), the conditional CDF at \(u\) is
\[
P(U \leq u \mid U \in(c, d))=\frac{P(U \leq u, c
Let \(U \sim \Unif{a,b}\). Claim: the distribution of \(X = c^{2} \log(d) U + e - f^{4}\) is still uniform when \(c,d, e, f \in \R^{+}\), \(c\neq 0\), \(d\neq 1\).
Solution
Solution, for real
True. This still is a linear transformation. Let \(c^{2} \log(d)\) be a constant \(c_{1}\). And let \(e - f^{4}\) be a constant \(c_{2}\). Then the question becomes \(c_{1} U + c_{2}\), this still is linear by BH.5.2.6.
Let \(Z \sim \Norm{0,1}\). Claim: \(\Phi(z) = \Phi(-z)\) due to symmetry.
Solution
Solution, for real
False. This equation for symmetry does not hold for the CDF. We have \(\phi(z) = \phi(-z)\). And \(\Phi(z) = 1 - \Phi(-z)\).
Let \(Z \sim \Norm{\mu, \sigma^{2}}\) with \(\sigma>0\) and \(X = Z \cdot \sigma^{-1} - \mu \cdot \sigma^{-1}\). Claim: \(X \sim \Norm{0,1}\).
Solution
Solution, for real
True. Rewriting results in \(X = \frac{Z - \mu}{\sigma}\). Then \(X \sim \Norm{0,1}\) by definition.
Let \(A\), \(B\) be two arbitrary events. Claim: \(\P{A \given B}=\P{A \cap B}/\P{B}\).
Solution
Solution, for real
False. The condition \(\P{B}>0\) is missing.
Let \(A\), \(B\) be events with \(\P{A},\P{B}>0\). Claim: \(\P{A \given B}\P{B}=\P{B \given A}\P{A}\).
Solution
Solution, for real
True.
Let \(A\), \(B\), \(C\) be events with \(\P{A \cap B}>0\). Claim:
\begin{equation*} \frac{\P{B \cap C \given A}}{\P{B \given A}}=\frac{\P{A \cap C \given B}}{\P{A \given B}}. \end{equation*}Solution
Solution, for real
True. Both sides are equal to \(\P{C \given A,B}\).
Let \(A\), \(B\), \(C\) be events with \(\P{C}>0\). Claim:
\begin{equation*} \P{A \cap B \given C} + \P{A \cup B \given C}=\P{A \given C}+\P{B \given C}. \end{equation*}Solution
Solution, for real
True.
Let \(A\) and \(B\) be two disjoint events with positive probability. Claim: \(A\) and \(B\) are dependent.
Solution
Solution, for real
True. \(\P{A \given B} = 0 \neq \P{A}\).
Suppose \(A_i\) for \(i=1, 2, \dots, n\) are independent indicator rvs. Claim: \(\sum_{i=1}^n A_i\) has a Binomial distribution.
Solution
Solution, for real
False. They also need to have identical distributions.
Let \(X\) and \(Y\) be independent. Claim: for any functions \(f, g\) it holds that \(g(X)\) is independent of \(f(Y)\).
Solution
Solution, for real
True.
Claim: a discrete rv requires that the number of outcomes is finite.
Solution
Solution, for real
False. It must have a countable number of outcomes, but not necessarily finite, like \(\N\).
We have an urn with \(w>0\) white balls and \(b>0\) black balls. We pick, without replacement, \(n\leq w+b\) balls. Then \(X \sim \HGeom{w, b, n}\). Claim: if \(X \sim \HGeom{3, 5, 2}\), then \(\P{X = 3} = 0\).
Solution
Solution, for real
True. 3 is not in the support of \(X\): \(\{0, 1, 2\}\).
Let \(c, d \in \R\), possibly the same. Take \(X\equiv c\) and \(Y\equiv d\). Claim: \(X\) and \(Y\) are independent.
Solution
Solution, for real
True. \(\P{X=c, Y=d} = 1 = \P{X=c}\P{Y=d}\), \(\P{X=c, Y\neq d} = 0 = \P{X=c} \P{Y\neq d}\), etc.
Let \(M_{X}(s)\) be the moment generating function of some rv \(X\). Claim: \(M_{X}(0) = 0\).
Solution
Solution, for real
False. Recall the definition: \(\E{e^{0\cdot X}} = \E{e^0} = 1\).
Let \(M_{X}(s)\) be the moment generating function of some rv \(X\). Claim:
\begin{equation*} \left(\frac{\d}{\d s}\right)^{2}M_{X}(s)\big|_{s=0} = \V{X} + (\E{X})^{2}. \end{equation*}Solution
Solution, for real
True.
We have two independent positive rvs \(X\) and \(Y\). Claim: \(M_{2X+Y}(s) = (M_{X}(s))^{2} M_{Y}(s)\).
Solution
Solution, for real
False. In general, because \(X\) is not independent of itself.
People enter a shop such that the time \(X\) between any two consecutive customers is \(X\sim \Exp{\lambda}\) with \(\lambda=10\) per hour. Claim: \(\P{X > x} = e^{-\lambda x}\), for \(x\geq 0\).
Solution
Solution, for real
True. See BH.5.45.
People enter a shop with iid interarrival times \(X\sim \Exp{\lambda}\). Let \(N(t)\) be the number of people during \([0,t]\). Claim: \(N(t) \sim \Pois{\lambda}\).
Solution
Solution, for real
False. See BH.5.45. It’s \(\sim \Pois{\lambda t}\).
With the same setup, suppose \(T_{3}\) is the time the third person enters. Claim: \(\P{N(t) <3} = \P{T_{3}>t}\).
Solution
Solution, for real
True.
Let \(X \sim \Exp{1}\) and \(Y = \lambda X\). Claim: \(Y \sim \Exp{\lambda}\).
Solution
Solution, for real
False. It should become \(Y \sim \Exp{\frac{1}{\lambda}}\).
Let \(X \sim \Exp{\lambda}\), then:
\begin{equation*} \int^{\infty}_{0} x \lambda e^{-\lambda x} \d x = \int^{\infty}_{-\infty} x \lambda e^{-\lambda x} \1{x> 0}\d x. \end{equation*}Solution
Solution, for real
True.
Let \(X\) be an exponential rv. Claim: \(\P{X > t + s \given X > t} = \P{X > t}\).
Solution
Solution, for real
False. It should be: \(\P{X > t + s \given X > s} = \P{X > t}\). This is the memoryless property.
Three cars in independent painting workshops, painting times iid \(X\sim \Exp{\lambda}\) with \(\E{X} = 1/\lambda = 3\). Claim: The expected time for the first two cars to be finished is 2.5 hours.
Solution
Solution, for real
True. Due to independence and memorylessness we have that the expected time for the first two cars to finish is \(T = T_{1} + T_{2}\), as we only need the first two painting jobs; \(T_{1}\) is the time for the first job to finish, \(T_{2}\) is the additional time for the second job to finish. Clearly \(T_{1} = \min\{X_{1}, X_{2}, X_{3}\}\), and, after a restart (recall, memoryless), \(T_{2} = \min\{X_{1}, X_{2}\}\). Hence,
\begin{equation*} \E{T} = \frac{1}{3 \lambda} + \frac{1}{2\lambda} = 1 + 1.5 = 2.5. \end{equation*}For more reference, see BH.5.6.3 and BH.5.6.5.
2. Week 2
For two strictly positive rvs \(X\) and \(Y\), let \(f_{X,Y}(x, y) = xy/(x^{2}+y^2)\). Claim: since
\begin{equation*} f_{X,Y}(x, y) = \frac{x}{\sqrt{x^{2}+y^2}} \frac{y}{\sqrt{x^{2}+y^2}}, \end{equation*}the rvs \(X\) and \(Y\) are independent.
Solution
Solution, for real
False. For independence of continuous rvs, the joint pdf must factor into two functions each depending on only one variable, i.e., \(f_{X,Y}(x,y) = f_{X}(x) f_{Y}(y)\). The two functions at the RHS of the claim are not of this type, both include \(x^{2}+y^{2}\).
The joint density of \(X\) and \(Y\) is \(f_{X,Y}(x,y) = Ce^{-(x + 2y)}\1{x\geq 0}\1{y\geq 0}\). Claim: \(X\) and \(Y\) are iid because \(f_{X,Y}(x, y) = C e^{-x}e^{-2y}\).
Solution
Solution, for real
False. The densities \(f_{X}\) and \(f_{Y}\) are not the same. Variations on this type of question: a. Yes, because … b. No, these rvs are identical, hence dependent. c. Yes, because \(X\) and \(Y\) are independent and identical. d. Yes, because \(X\) and \(Y\) are independent and identically distributed.
Claim: \(p(k) = \frac{1-x}{1-x^{n+1}} x^k\) with \(k \in \{0, \dots, n\}\) is a valid PMF for any \(x\).
Solution
Solution, for real
False. Not if \(x = 0\) or \(x = 1\).
Let \(X \sim \Geo{p}\). Claim: all of the following steps are correct.
\begin{equation*} \P{X = k \given X \geq n} = \frac{\P{X=k, X \geq n}}{\P{X \geq n}} = \frac{\1{k \geq n} \P{X = k}}{\P{X \geq n}} = \1{k \geq n} pq^{k-n} \end{equation*}Thus,
\begin{align*} \E{X \given X \geq n} &= \sum_{k=0}^\infty k \1{k \geq n} pq^{k-n}\\ &= p \sum_{k=n}^\infty kq^{k-n}\\ &= (1 - q) \big(n q^0 + (n+1) q^1 + (n+2) q^2 + \dots\big)\\ &= \big(n q^0 + (n+1) q^1 + (n+2) q^2 + \dots\big) - \big(n q^1 + (n+1)q^2 + (n+2)q^3 + \dots\big)\\ &= n + q^1 + q^2 + \dots\\ &= n + \frac{1}{1-q} - 1\\ &= n + \frac{q}{p}\\ \end{align*}Solution
Solution, for real
True. Of course, the memoryless property of the geometric distribution can be used to find the answer too.
Claim: suppose \(f(x) = a x + b\) with \(a\neq 0\), then there is a \(c\) such that \(c e^{-(f(x))^2}\) is the pdf of a normal distribution.
Solution
Solution, for real
True.
Let \(X\sim \Geo{p}\). Take \(s\) such that \(e^{s}q < 1\).
\begin{align*} M_{X}(s) &\stackrel{1}= \E{e^{sX}} \stackrel{2}= \sum_{k=1}^{\infty} p q^{k} e^{sk} \stackrel{3}= p \sum_{k=1}^{\infty} (e^{s}q)^{k} \stackrel{4}= p \left(\sum_{k=0}^{\infty} (e^{s}q)^{k}-1\right) \stackrel{5}= \frac{p}{1+e^{s}q} - p. \end{align*}Claim: more than one of these steps is incorrect.
Solution
Solution, for real
True. Steps 2 and 5 are incorrect. Step 2: start with \(k=0\), step 5: the plus should be a minus. Variations on this theme: a. all steps are correct. b. step 2 is incorrect.
Claim: \(M(t) = 2^{-n} \sum_{k=0}^n \binom{n}{k} e^{tk}\) is a valid expression for the MGF of a \(\Bin{n, 0.5}\) distribution.
Solution
Solution, for real
True. Apply the binomial theorem: \((e^t + 1)^n = \sum_{k=0}^n \binom{n}{k} e^{tk}\).
Claim: \(M_X(s)=e^{-(s-1)^2/2}\) could be a valid MGF for some rv \(X\).
Solution
Solution, for real
False. Note that \(M_X(s)=\E{e^{sX}}\), s.t. \(M_X(0)=1\) must always hold.
Claim: this contains an error:
\begin{equation*} e \stackrel{1}= \lim_{n\to\infty} (1+n^{-1})^n \stackrel{2}= \lim_{n\to\infty} \sum^n_{k=0} n^{-k}. \end{equation*}Solution
Solution, for real
True. Step 2 is wrong. Compare the Taylor series for \(e^{x}\).
Let \(X\) and \(Y\) be two independent rvs such that \(\E{e^{s X}}\) and \(\E{e^{s Y}}\) are well defined for \(s\) in an open interval around \(0\). Claim: \(M_{X-Y}(s)=M_X(s)M_Y(-s)\).
Solution
Solution, for real
True. \(M_{X-Y}(s)=\E{e^{s(X-Y)}}=\E{e^{sX}}\E{e^{-sY}}=M_X(s)M_Y(-s)\).
Let \(X\sim \Pois{\lambda}\) and \(Y \sim \Pois{\mu}\) be independent. Claim: \(M_{X+Y}(t) = e^{(\lambda \mu)(e^t-1)}\).
Solution
Solution, for real
False. By Taylor series we find \(\E{e^{tX}} = e^{-\lambda} e^{\lambda e^t}\) and by independence \(M_{X+Y}(t) = M_X(t) \cdot M_Y(t) = e^{\lambda(e^t-1)} e^{\mu(e^t-1)} = e^{(\mu + \lambda)(e^t-1)}\)
Let \(X_1\sim \Norm{\mu_1, \sigma_1^2}\) and \(X_2\sim \Norm{\mu_2, \sigma_2^2}\). Claim: \(M_{X_1+X_2}(t) = e^{(\mu_1 + \mu_2)t +\frac{1}{2}(\sigma_1^2 + \sigma_2^2)t^2}\).
Solution
Solution, for real
False. It is not given that \(X_1\) and \(X_2\) are independent.
3. Week 3
Let \(X \sim \Norm{0,1}\), \(Y=X^2\). Claim: for the density of \(Y\), \(f_{Y}(0) = 0\), and on \(y>0\),
\begin{equation*} f_Y(y)= \frac{\phi(\sqrt{y}) + \phi(-\sqrt{y})}{2\sqrt{y}} \end{equation*}by the change of variables formula.
Solution
Solution, for real
True. Consider the set \(A = \{x : x^{2} =y\}\) as the inverse of \(y\). The change of variables formula says this
\begin{equation*} f_Y(y)= \sum_{x_{i} \in A} f_{X}(x_{i}) \left(\frac{\d y}{\d x}(x_{i})\right)^{-1}, \end{equation*}if \(\d y/\d x (x_{i}) \neq 0\) for all \(x_{i} \in A\). As it is given that \(y>0\), this condition is satisfied.
Let \(X \sim \Norm{0,1}\), \(Y=X^2\). Claim: The change of variables formula tells us that:
\begin{equation*} f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|=\phi(x)\left|\frac{1}{2x}\right|=\phi(\sqrt{y})\frac{1}{2\sqrt{y}},\quad y>0. \end{equation*}Solution
Solution, for real
False. \(Y=X^2\) is not a strictly increasing (or decreasing) function, so the change of variables formula cannot be applied.
Let \(X \sim \Gamm{n,\lambda}\). Then \(\E{X^{k}} = \frac{n+k-1}{\lambda} \E{X^{k-1}}\) for \(k \in \N\). Claim: for \(c\in \N\), \(\E{X^{c}} = (n+c-1)!/((n-1)!\lambda^{c})\).
Solution
Solution, for real
True.
Let \(X\) be a discrete rv on \(\{a_{i}\}_{i=1}^{\infty}\) with \(a_{i} \in \R\). Claim: the PMF of \(X\) can be found by \(f_{X}(x) = F_{X}'(x)\) for \(x\in \{a_{i}\}\).
Solution
Solution, for real
False. The PMF does not have a derivative (in the proper sense) at \(a_{i}\).
Suppose the rv \(X\) has PDF \(f_{X}(x) = Ax^{-s} \1{x\geq 1}\) for \(s\in (1, 2)\) where \(A\) is the normalization constant. Claim: \(\E{X} = \infty\).
Solution
Solution, for real
True.
Claim: LOTP on a discrete sample space \(S\) states that \(\P{B} = \sum_{i=1}^{n} \P{B \given A_{i}} \P{A_{i}}\), where \(\{A_{i}\}\) is a set of non-overlapping subsets of \(S\).
Solution
Solution, for real
False. In general, because it’s not given that the subsets cover \(S\).
Let \(X\) be a rv on \(\R\) and \(g\) a function from \(\R^{2}\) to \(\R\). Claim: \(g(X)\) is a rv.
Solution
Solution, for real
False. \(g\) needs two arguments as it maps \(\R^{2}\) to \(\R\).
Let \(X\sim \FS{p}\) with \(p\in (0, 1)\), \(q=1-p\). Claim: \(\E{X} = 1+q\E{X} \implies \E{X} = 1/p\).
Solution
Solution, for real
True.
Claim: according to 2D LOTUS, if \(g(x, y) \in \R\) and \(X, Y\) are real-valued rvs, then
\begin{equation*} \E{g(X,Y)} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x,y) f_{X}(x)f_Y(y) \d x \d y. \end{equation*}Solution
Solution, for real
False. \(X\) and \(Y\) need not be independent, as is suggested here.
Let \(L = \min\{X_{i} : i =1, \ldots, n\}\) with \(\{X_{i}\}\) iid. Claim:
\begin{equation*} \P{L\leq x} = (\P{X_{1} \leq x})^{n}. \end{equation*}Solution
Solution, for real
False. This holds for the maximum, or reverse the direction of the inequalities with respect to \(x\).
Let \(X, Y\) be two continuous rvs with joint distribution \(F_{X,Y}(x,y)\). Claim: \(f_{X}(x) = \partial_{x} F_{X,Y}(x,y)\).
Solution
Solution, for real
False. Why is this claim nonsense?
Claim: for two continuous rvs \(X, Y\) with joint PDF \(f_{X,Y}(x,y)\), \(f_{Y \mid X}(y \mid x) = F_{X,Y}(x,y)/F_{X}(x)\).
Solution
Solution, for real
False.
Let \(X\) and \(Y\) be independent discrete rvs, \(T = X + Y\). Claim: \(F_T(t) = \sum_x F_X(x - t) p_X(x)\).
Solution
Solution, for real
False. It should be \(F_Y(t - x)\). Variations:
- True
- False with \(p_X(t)\)
- False with \(\sum_x p_Y(t - x) p_X(x)\)
- False with \(\sum_{x=0}^t F_Y(t - x) p_X(x)\)
Let \(X_1, \cdots, X_n\) be iid continuous rvs with CDF \(F\), and \(N_{x} \sim \Bin{n,F(x)}\). Claim: The CDF of the \(j\)th order statistic can be written as \(\P{X_{(j)} \leq x} = \P{N_{x} \geq j}\).
Solution
Solution, for real
True. See BH. 400
4. Week 4
You may assume that, when \(T=X+Y\), \(f_{X,T}(x,t) = f_{X,Y}(x, t-x)\), even when the positive rvs \(X\) and \(Y\) are not independent. Claim: \(f_T(t) = \int_{0}^{t} f_{X, Y}(x,t-x) \d x\).
Solution
Solution, for real
True.
You may assume that, when \(T=X+Y\), \(f_{X,T}(x,t) = f_{X,Y}(x, t-x)\). Claim: When the rvs \(X, Y\) have support equal to \(\R\) and are independent,
\begin{equation*} f_T(t) = \int_{0}^{t} f_{Y}(t-x) f_X(x) \d x. \end{equation*}Solution
Solution, for real
False. It is not given that \(X\) or \(Y\) are positive.
The positive rvs \(X\) and \(Y\) are independent, and \(T=XY\). Claim: \(f_T(t) = \int_{0}^{t} f_{Y}(t/x) f_X(x) \d x\).
Solution
Solution, for real
False. The integration bounds are incorrect, which is easy to see. The harder part is that the Jacobian is missing. Even without knowing this, one can guess from the 1D case that there must be a function to compensate for the change in \(f_{Y}\), since we divide \(t\) by \(x\).
Let \(X_{(i)}\), \(i = 1, \ldots, n\) be the order statistic of \(\{X_{i}\}\). Claim:
\begin{equation*} \P{X_{(j)}\leq x} = \sum_{k=0}^{n}\binom{n}{k} F(x)^k (1-F(x))^{n-k}. \end{equation*}Solution
Solution, for real
False. Check theorem 8.6.3. It’s easy to check. The LHS depends on \(j\), the RHS not, hence it must be false, unless the probability would not depend on \(j\), but that cannot be true.
Let \(X_{(i)}\), \(i = 1, \ldots, n\) be the order statistic of continuous rvs \(\{X_{i}\}\). Claim:
\begin{equation*} f_{(j)}(x) \d x = n f(x) \d x \binom{n}{j} F(x)^j (1-F(x))^{n-j}. \end{equation*}Solution
Solution, for real
False. Theorem BH.8.6.4.
Claim: It is a good idea to conceptualize the order statistic as a set rather than as a list.
Solution
Solution, for real
False. Elements in a set are not ordered, elements in a sequence or list are.
Let \(X_{(i)}\), \(i = 1, \ldots, n\) be the order statistic of continuous iid rvs \(\{X_{i}\}\). Claim: \(\E{X_{(i)} \given X_{(j)}=x} \leq x\) for any \(i\leq j\).
Solution
Solution, for real
True.
Claim: \(\E{h(Y)Y \given X} = h(Y) \E{Y \given X}\).
Solution
Solution, for real
False.
Claim: \(\E{h(Y) \given Y} = h(Y)\E{1 \given Y} = h(Y)\cdot 1 = h(Y)\).
Solution
Solution, for real
True.
Let \(g(x) =\E{Y \given X=x}\). Claim: the following rv
\begin{equation*} g(X) = \sum_{y=-\infty}^{\infty} y \P{Y=y \given X=X} \end{equation*}is well-defined.
Solution
Solution, for real
False. The final condition \(X=X\) is nonsense.
Claim: \(\V{X \given Y} = \E{X^2 \given Y} - (\E{X \given Y})^{2}\).
Solution
Solution, for real
True.
Claim: \(\V{\E{X \given Y}} = \E{X^2 \given Y} - (\E{X \given Y})^{2}\).
Solution
Solution, for real
False. The first term of the RHS should be \(\E{(\E{X \given Y})^{2}}\). The second term is also wrong: it must be \((\E{Y})^{2}\). Another way to see why the claim is false is like this. The LHS is the variance of the rv \(\E{X \given Y}\). Thus, the LHS is some constant (probably \(>0\)). The RHS still depends on \(Y\), hence it is a rv.
We have two rvs \(X\) and \(Y\), such that \(X\) is independent of \(Y\). Claim: this does not imply that \(Y\) is independent of \(X\).
Solution
Solution, for real
False. Independence is symmetric.
Claim: if \(X\) is independent from \(Y\), then \(\E{\V{Y \given X}} = \V{Y}\).
Solution
Solution, for real
True. \(\V{Y \given X}\) is a rv, whereas \(\V{Y}\) is a number. Taking the expectation of \(\V{Y \given X}\) reduces the rv to a number. The book writes, as a property, that \(\E{Y \given X} = \E{Y}\) when \(Y,X\) are independent. We use the same property here for the variance.
Let \(N \sim \Pois{\lambda}\), \(X_j\) iid with mean \(\mu\). Claim: \(\E{\sum_{j=1}^N X_j}=\lambda\mu\).
Solution
Solution, for real
True.
Claim: \(\E{Z \mid \E{X \mid Y}}\) is a non-degenerate rv and a function of \(Y\).
Solution
Solution, for real
False. It is not given that \(X\), \(Y\) and \(Z\) are dependent.
Claim: \(\E{\E{Y \given X,Z} \given Z} = \E{Y \given X}\).
Solution
Solution, for real
False. By definition: \(\E{\E{Y \given X,Z} \given Z} = \E{Y \given Z}\).
Let \(X\) and \(Y\) be any rvs. Claim: all of the following steps are correct:
\begin{align*} \V{Y-\E{Y \given X}} &= \E{(Y-\E{Y \given X})^2} - (\E{Y-\E{Y \given X}})^2 \\ &= \E{(Y-\E{Y \given X})^2} - \E{\E{(Y-\E{Y \given X})^2 \given X}} \\ &= \E{\V{Y \given X}}. \end{align*}Solution
Solution, for real
True. See BH 435
5. Week 6
Claim: \(\E{Y \given A} = \sum_{y=0}^{\infty} y \P{Y=y \given A}\) if \(Y\) is discrete.
Solution
Solution, for real
False. What if \(Y\) can take negative values?
Claim: \(\E{Y \given A} = \sum_{y=-\infty}^{\infty} y \P{Y=y \given A} = 0\) if \(\P{A}=0\) and \(Y\) is discrete.
Solution
Solution, for real
False. The definition in BH requires that \(\P{A}>0\).
Let \(X\) be a continuous rv with PDF \(f_{X}(x) > 0\) on \(x\in[a, b]\) and \(Y\) discrete. Claim: when \(x\in [a, b]\),
\begin{equation*} \E{Y \given X=x} = \sum_{y=-\infty}^{\infty} y \frac{f_{X,Y}(x,y)}{f_{X}(x)}. \end{equation*}Solution
Solution, for real
True.
Take \(g(x) = \E{Y \given X=x}\). Define the conditional expectation of \(Y\) given \(X\) as \(g(X)\), and write it as \(\E{Y \given X}\). Claim: this is one of the most important definitions in probability.
Solution
Solution, for real
True. This is just a bonus to stress the importance of the definition of conditional expectation.
Let \(X\) be a continuous rv with support \((0, \infty)\). Claim: for \(x>0\), \(\E{X \given X>x}>\E{X}\).
Solution
Solution, for real
True.
Let \(X \sim \Exp{\lambda}\); we write \(f_{X}\) for the density of \(X\). Claim: all steps in the following lines are correct.
\begin{align*} f_{X}\rb{x \given X>s} &= \frac{\1{x > s} \lambda e^{-\lambda x}}{e^{-\lambda s}} \\ \E{X \given X>s} &= \int^{\infty}_{s} x \lambda e^{-\lambda(x-s)} \d x = \int^{\infty}_{0}(x + s) \lambda e^{-\lambda x} \d x. \end{align*}Solution
Solution, for real
True.
Let \(X\) and \(Y\) be two rvs. Claim: if all is well defined and finite, \(\E{X \given Y} = c\), where \(c\) is some constant.
Solution
Solution, for real
False. \(\E{X \given Y}\) is a function of \(Y\), which is a rv.
Let \(X\) be a rv and \(A\) an event. Claim: \(\E{X \mid \1{A}} = \E{X \given A}\).
Solution
Solution, for real
False. One way to see this is to note that the LHS is a rv, while the RHS is a number. In fact, \(\E{X \mid \1{A} = 1} = \E{X \given A}\). An alternative question: Is \(\E{X\1{A}} = \E{X \given A} \P{A}\)?
The continuous rvs \(\{X_{i}\}\) with support on \((0, \infty)\) are iid, and \(S_n=\sum_{i=1}^n X_{i}\). Claim: for some \(x\) and \(n\),
\begin{equation*} \E{X_{n} \given S_{n-1} = x} = S_n/n. \end{equation*}Solution
Solution, for real
False. The condition is on some given \(x\), so on the LHS we have an \(x\). However, the RHS contains no \(x\). Moreover, we condition on \(S_{n-1}\), which depends on \(X_{1}, X_{2}, \ldots, X_{n-1}\), but \(X_{n}\) is independent of these rvs.
6. Week 7
Eve’s law says that \(\V{X} = \E{\V{X \given N}} + \V{\E{X \given N}}\). Claim: \(\E{\V{X \given N}}\) is called the in-between group variation.
Solution
Solution, for real
False.
Eve’s law says that \(\V{X} = \E{\V{X \given N}} + \V{\E{X \given N}}\). Claim: \(\V{\E{X \given N}}\) is called the explained variance.
Solution
Solution, for real
True.
Write \(g(X) = \E{Y \given X}\) for two rvs \(X, Y\). Claim: This derivation is correct:
\begin{equation*} \V{\E{Y \given X}} = \E{(g(X))^2} - (\E{g(X)})^{2} = \E{(g(X))^2} - (\E{Y})^{2}. \end{equation*}Solution
Solution, for real
True.
Claim: The inequality of Cauchy-Schwarz says that \(\E{(XY)^{2}} \geq \E{X} \E{Y}\).
Solution
Solution, for real
False.
Claim: This is correct: \(\E{X} \leq \E{X\1{X\geq 0}}\).
Solution
Solution, for real
True.
Let \(g\) be a function that is concave and convex at the same time, and such that \(g(0) = 0\). Claim: by Jensen’s inequality: \(\E{g(X)} = g(\E{X})\).
Solution
Solution, for real
True.
Claim: The following reasoning is correct. For any \(a\geq 0\),
\begin{equation*} \E{|X|} \geq \E{a \1{|X|\geq a}} = a \P{|X| \geq a}. \end{equation*}Solution
Solution, for real
True.
The set \(\{X_i : i = 1, 2, \ldots\}\) forms a set of iid rvs such that \(X_i\in \{0, 1\}\) and \(\P{X_i=1} = 1/2\) for all \(i\). Take \(A=\{\lim_{n\to\infty} n^{-1}\sum_{i=1}^{n}X_i = 1\}\). Claim: The strong law of large numbers implies that \(A=\varnothing\).
Solution
Solution, for real
False.
The set \(\{X_i : i = 1, 2, \ldots\}\) forms a set of iid rvs such that \(X_i\in \{0, 1\}\) and \(\P{X_i=1} = 1/2\) for all \(i\). Take \(A=\{\lim_{n\to\infty} n^{-1}\sum_{i=1}^{n}X_i = 1\}\). Claim: The strong law of large numbers says that \(\P{A} = 0\).
Solution
Solution, for real
True.
The set \(\{X_i : i = 1, 2, \ldots\}\) forms a set of iid rvs. Claim: The weak law of large numbers states that:
\begin{equation*} \forall \delta, \epsilon > 0: \exists m > 0 : \forall n > m: \P{|\bar X_{n}-\mu| > \epsilon} < \delta, \end{equation*}where \(\mu=\E{X_{i}}\) and \(\bar X_n = n^{-1}\sum_{i=1}^{n}X_{i}\).
Solution
Solution, for real
True.
Let \(X_i \sim \Exp{\lambda}\) with \(Y_i = 2X_i\), \(i=1,2,\dots\). Claim: \(\P{\bar{X}_n\bar{Y}_n \to 2/\lambda^2}=1\).
Solution
Solution, for real
True.
Claim: the Chernoff bound is always tighter than the Chebyshev bound and both are always tighter than the Markov bound.
Solution
Solution, for real
False.
Claim: the equation \(\E{|X|} = |\E{X}|\) only holds for rvs with a strictly positive support.
Solution
Solution, for real
False. Every rv with a nonnegative support satisfies this equation, zero included.
Let \(X_1, X_2, \cdots\) be iid fair coin tosses. Let \(\bar{X}_n\) be the fraction of heads after \(n\) tosses. Claim: By SLLN \(\bar X_n \to 1/2\) as \(n\to \infty\) with probability 1.
Solution
Solution, for real
True.
7. Week 8
Let \(X_i\) be rvs with mean \(\mu\) and variance \(\sigma^2\), and \(\bar X_n = n^{-1}\sum_{i=1}^n X_i\). Claim: the CLT states that \(\bar X_n \stackrel{\cdot}\sim \Norm{\mu, \sigma^2 \sqrt{n}}\).
Solution
Solution, for real
False. First, the \(X_i\)’s need to be independent. Second, the variance of \(\bar X_{n}\) should be \(\sigma^2/n\).
Let \(\{X_{n}\}\) be a sequence of iid rvs with an unknown distribution but with finite mean \(\mu\) and variance \(\sigma^{2}\). If \(\bar{X}_{n} \to \Norm{\mu, \sigma^{2}/n}\) as \(n \to \infty\) then \(X_{n}\) must be normally distributed for all \(n\).
Solution
Solution, for real
False. Given a sequence of iid rvs \(\{X_{n}\}\) with finite mean and variance, \(\bar{X}_{n} \to \Norm{\mu, \sigma^{2}/n}\) as \(n \to \infty\) by the CLT. This holds for all such sequences of rvs. Not only the normal distribution.
Let \(X_{i} \sim \Pois{\lambda}\) be iid. Claim:
\begin{equation*} \left( \frac{ \sum^{n}_{i=1} X_{i}}{\sqrt{n} \sqrt{\lambda}} - \sqrt{n}\sqrt{\lambda} \right) \to \Norm{0,1} \quad \text{as } n \to \infty. \end{equation*}Solution
Solution, for real
True. \[\sqrt{n} \left( \frac{ \bar{X}_{n} - \lambda}{\sqrt{\lambda}} \right)\] \[\sqrt{n} \left( \frac{ \frac{1}{n} \sum^{n}_{i=1} X_{i} }{\sqrt{\lambda}} - \sqrt{\lambda} \right)\] \[ = \left( \frac{ \sum^{n}_{i=1} X_{i}}{\sqrt{n}\lambda} - \sqrt{n}\sqrt{\lambda} \right)\] \[ \to \Norm{0,1} \quad \text{as } n \to \infty\]
Let \(V \sim \chi^2_n\). Claim: For large \(n\), \(V \stackrel{\cdot}\sim \Norm{n, 2n}\).
Solution
Solution, for real
True. Check the definition.
Let \(T_n = Z/\sqrt{V_n/n}\), with \(Z\sim \Norm{0,1}\), \(V_n\sim \chi_n^2\), \(Z\) and \(V_n\) independent. Claim: because by SLLN \(V_n/n \to \E{Z_1^2}\) with probability 1, \(T_n\) approaches the standard normal distribution.
Solution
Solution, for real
True. See BH 480
Let \(Z_1^2 + \cdots + Z_n^2 = V \sim \chi^2_n\), where \(Z_1, \ldots, Z_n \sim \Norm{0,1}\) are iid, and given \(\E{Z_1^4} = 3\). Claim: \(\V{V} = 2n^2\).
Solution
Solution, for real
False. \(\V{V} = n\V{Z_1^2} = n\left(\E{Z_1^4} - \E{Z_1^2}^2\right) = 2n\).
Let \(\bar{Z}_n=n^{-1}\sum_{i=1}^n Z_i\) where \(Z_i\sim\Norm{0,1}\), \(i=1,\dots,n\). Claim: \(\bar{Z}_n\sim\Norm{0, 1/n}\), so that \(n\bar{Z}_n^2\sim\chi^2_1\).
Solution
Solution, for real
True. \(\sqrt{n}\bar{Z}_n\sim\Norm{0,1} \implies n\bar{Z}_n^2\sim\chi^2_1\).
Let \(Z_i \sim \Norm{0,1}\), \(i=1,\dots,n\) with \(V_n=\sum_{i=1}^n Z_i^2\). Claim: \(Z_1/\sqrt{V_n/n} \sim t_n\).
Solution
Solution, for real
False. \(Z_1\) and \(V_n\) are dependent.